The required functions are $f(x) = a(x-b)^n$, $0 \le x \le 1$, where $n$ is an integer greater than $1$, and $a$ and $b$ are rational numbers, $a \neq 0$ and $b \notin [0, 1]$. Clearly, these functions satisfy the conditions in the statement.

Since the closed unit interval $[0, 1]$ is uncountable, and there are only countably many polynomial functions with rational coefficients, there exist an uncountable set $S \subseteq [0, 1]$ and coprime polynomial functions $g, h: [0, 1] \to \mathbb{R}$ with rational coefficients such that $h(x) \neq 0$ and $\int_0^x \frac{1}{f(t)} dt = \frac{g(x)}{h(x)}$, for all $x$ in $S$. Since at most countably many points of $S$ are not accumulation points of $S$, and $S$ is uncountable, it follows that $S$ contains infinitely many of its accumulation points. At each of these points, the rational functions $1/f$ and $(g/h)' = (g'h - gh')/h^2$ are equal, so
$$h^2 = f \cdot (g'h - gh') \quad (*)$$
in $\mathbb{Q}[X]$. Since $\deg f \ge 1$, it follows that $\deg g \le \deg h$. Notice that the remainder of $g$ upon division by $h$ also satisfies (*) to assume henceforth $\deg g < \deg h$, so $\deg (g'h - gh') = \deg g + \deg h - 1$.

If $\deg h = 1$, then $\deg g = 0$ and $f = a(X - b)^2$, where $a$ and $b$ are rational numbers, $a \neq 0$ and $b \notin [0, 1]$.

If $\deg h \ge 2$, since $g$ and $h$ are coprime, (*) implies that every $k$-fold root (not necessarily real) of $g'h - gh'$ is a $(k+1)$-fold root of $h$, so $\deg h \ge \deg g + \deg h - 1 + d$, where $d$ is the number of distinct roots of $g'h - gh'$. Consequently, $d = 1$ and $\deg g = 0$, so $g$ is a non-zero constant, $h = c(X - b)^n$, where $b$ and $c$ are rational, $b \notin [0, 1]$, $c \neq 0$, and $n$ is an integer greater than $1$, and $f = a(X - b)^{n+1}$ for some non-zero rational $a$.